Given 4 GB (≈ 4.3 × 10⁹ bytes) of virtual address space and a page size of 4 KB (= 4 × 10³ bytes), with each page table entry 5 bytes: how many virtual pages would this imply, and what is the size of the whole page table?
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Correct answer: D. None of the above
Explanation
Number of virtual pages = virtual space ÷ page size = (4.3×10⁹) ÷ (4×10³) = 1,075,000 pages. Page table size = number of entries × entry size = 1,075,000 × 5 bytes = 5,375,000 bytes (≈ 5.375 MB, i.e. 5375 KB). None of the three numeric options printed on the source page (107500/20480, 215000/40960, or 10750/10240) match this calculation, however it is scaled — they appear to be garbled figures from an unrelated version of the question. This exact NTA question is independently documented (e.g. on Testbook) as having been officially dropped by UGC/NTA for this reason, with the recorded correct choice being 'None of the above'.More Programming Languages and Computer Graphics questions
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